Poset fiber theorems under a special assumption on the poset map?!

Question

Hey everyone, I am facing the following problem:

Say that a (order-preserving) poset map $f:P\to Q$ has property $(\star)$ if for all $q_1,q_2\in Q$ with $q_1\leq q_2$ and every $p_2\in f^{-1}(q_2)$ there exists an element $p_1\in f^{-1}(q_1)$ such that $p_1\leq p_2$.

Given such an $f$, can we already make any statements about its geometric realization $|f|:|P|\to |Q|$, which is a continuous map between topological spaces? I am especially interested in anything that relates the homotopy types of $|P|$ and $|Q|$, as in my explicit case I know that $|Q|$ is contractible and want to show that this topological property also holds for $|P|$.

In my case, $f$ is surjective and for all $q\in Q$ the simplicial complex $|f^{-1}(q)|$ is homeomorphic to a tree, thus contractible. The Quillen fiber lemma states that the contractibility of $|f^{-1}(Q_{\leq q})|$ for all $q\in Q$ already implies that $|f|$ is a homotopy equivalence, and now I am interested in some modified version of the Quillen fiber lemma which tells me that if $f$ has property $(\star)$, then it suffices to check contractibility on the subcomplexes $|f^{-1}(q)|\subseteq |f^{-1}(Q_{\leq q})|$.

I have thought about several counterexamples where it does not suffice to check contractibility on the (geometrically realized) fibers over single points in $Q$, but all of the counterexamples' poset maps did not have property $(\star)$.

Thanks a ton in advance.

Sebastian

Answer 1

Unfortunately, property $(\star)$ is not enough. Here is an easy counterexample.

Answer 2

In the following, I shall assume that the posets $P$ and $Q$ are finite.

Then it is at least true from the condition that $f^{-1}(q)$ is contractible for all $q \in Q$ that the map $|f|$ is a homology isomorphism, by the Vietoris-Begle theorem (http://en.wikipedia.org/wiki/Vietoris%E2%80%93Begle_mapping_theorem). This is a very old result.

(Notice: I do not even require your condition ($\ast$))

In fact it is even true that your map is a homotopy equivalence (under the assumption that point inverses are contractible) since $|f|$ is a simplicial map of simplicial complexes which has contractible point inverses. This is what people call a "simple map" in simple homotopy theory. It's a basic result to the subject that a simple map of poyhedra is a homotopy equivalence. This result dates from the late 1960s I think, perhaps the 1970s.

Addendum: here's a reference:

Chapman, T. A. Cell-like mappings. Algebraic and geometrical methods in topology (Conf. Topological Methods in Algebraic Topology, State Univ. New York, Binghamton, N.Y., 1973), pp. 230–240. Lecture Notes in Math., Vol. 428, Springer, Berlin, 1974.

Second Addendum: What I wrote above is pure bunk. We also need to know that $|f|^{-1}(x)$ is constractible for all points $x$---not just the vertices. Here's what I think to be the case: Suppose one has the additional condition that $f^{-1}(\sigma)$ is contractible, where $\sigma = x_0 \le x_1 \le \cdots \le x_k$ is any finite chain. Then it looks to me as if the additional condition will guarantee that $|f|$ is a homotopy equivalence.

Now it seems to me that your condition ($\ast$) amounts to no more than the statement that $f^{-1}(x_0 \le x_1)$ is non-empty. So my condition will imply yours, but not vice-versa.